# If $M=\sum_{g\in G}\rho(g)$ then $\operatorname{tr}(M)=0$ implies $M=0$

Let G be a finite group and $$\rho:G\to GL_n(\mathbb C)$$ a representation.

a) If $$M=\sum_{g\in G} \rho (g) \neq 0$$ then prove that there is a non-zero vector $$v$$ such that $$\rho(g)v=v$$ for every $$g\in G$$.

b) If $$\sum_{g\in G} \chi(g)=0$$ then prove that $$M=0$$

The first part is easy, as if $$M\neq 0$$, then there s a vector $$w$$ such that $$Mw=v \neq 0$$. However, then $$\rho(g)v= \rho(g) M w= Mw=v$$ for every g Any hint would be welcome.

For b) however I am a bit stuck. $$\sum_{g\in G} \chi(g)=\operatorname{tr}(M)$$. Suppose that $$M\neq 0$$, and consider v be the vector found in (a), then if I complete $$v$$ to a basis of $$\mathbb C^n$$ then the first column of every matrix $$\rho(g)$$ would be the vector $$[1,0,...,0]$$. However, on the rest of the diagonal I could have negative values so I dont see why trace being $$0$$ implies $$M=0$$

• Hint: Show that $1/\left|G\right|\cdot M$ is idempotent. What do you know about the trace of an idempotent matrix? In particular, how is it connected to the rank of said matrix? Jul 5 '19 at 17:25

Notice that $$\sum_{g\in G} \chi(g)$$ is the inner product of the representation with the trivial representation, hence if it equals $$0$$ it means that the decomposition in irreducible representation does not contain the trivial representation. But if $$M\neq 0$$ then $$\rho$$ has a one dimensional invariant subspace, i.e. a trivial subrepresentation, thus $$M=0$$.